重积分
XantC
二重积分
定义
$$ \iint_{D}f(x,y)\mathrm{d}\sigma = \lim_{\lambda \to 0}\sum_{i=1}^{n}f(\xi_i, \eta_i)\Delta \sigma_i $$
计算
直角坐标
$$ \begin{align*} \iint_{d}f(x,y)\mathrm{d}\sigma&=\int^b_a\mathrm{d}x\int^{\varphi_1(x)}_{\varphi_2(x)}f(x,y)\mathrm{d}y\\ &=\int^d_c\mathrm{d}y\int^{\psi_1(y)}_{\psi_2(y)}f(x,y)\mathrm{d}x \end{align*} $$
极坐标
ρ1 ≤ ρ ≤ ρ2, α ≤ θ ≤ β
$$ \begin{align*} \iint_Df(\rho\cos\theta, \rho\sin\theta)\rho\mathrm{d}\rho\mathrm{d}\theta=\int^\beta_\alpha\mathrm{d}\theta\int^{\rho_2}_{\rho_1}f(\rho\cos\theta, \rho\sin\theta)\rho\mathrm{d}\rho \end{align*} $$
求高斯函数
对于D = {(x, y)|x2 + y2 ≤ a2}:
∬De−x2 − y2 = ∬D′e−ρ2ρdρdθ = π(1 − e−a2)
令D1 = {(x, y)|x2 + y2 ≤ R2, x ≥ 0, y ≥ 0}
$$ D_2=\{(x,y)|x^2+y^2\leq(\sqrt{2}R)^2,x\geq0,y\geq0\} $$
S = {(x, y)|0 ≤ x ≤ R, 0 ≤ y ≤ R}
⇒ ∬D1e−x2 − y2dxdy < ∬Se−x2 − y2dxdy < ∬D2e−x2 − y2dxdy
$$ \begin{align*} \because &\iint_{D_1}e^{-x^2-y^2}\mathrm{d}x\mathrm{d}y=\frac{\pi}{4}(1-e^{-R^2})\\ &\iint_{S}e^{-x^2-y^2}\mathrm{d}x\mathrm{d}y=\int^R_0 e^{-x^2}\mathrm{d}x\cdot\int^R_0 e^{-y^2}\mathrm{d}y=(\int^R_0 e^{-x^2}\mathrm{d}x)^2\\ &\iint_{D_2}e^{-x^2-y^2}\mathrm{d}x\mathrm{d}y=\frac{\pi}{4}(1-e^{-2R^2}) \end{align*} $$
$$ \therefore\frac{\pi}{4}(1-e^{-R^2})<(\int^R_0 e^{-x^2}\mathrm{d}x)^2<\frac{\pi}{4}(1-e^{-2R^2}) $$
$$ R\to +\infty\colon \int^{+\infty}_0e^{-x^2}=\frac{\sqrt{\pi}}{2}$$
一般换元
$$ x=x(u,v),\;y=y(u,v),\;J(u,v)=\frac{\partial(x,y)}{\partial(u,v)} $$
∬Df(x, y)dxdy = ∬D′f(x(u, v), y(u, v))|J(u, v)|dudv
证明
证明的关键是表示面积元素
在u − v坐标系取一小正方形M1′M2′M3′M4′, 则在x − y坐标系中的对应曲变形为M1M2M3M4
令M1′(u, v), M2′(u + h, v), M3′(u + h, v + h), M4′(u, v + h)
M1: x1 = x(u, v), y1 = y(u, v)
$$ \begin{align*} M_2\colon&x_2=x(u+h,v)=x(u,v)+x_u(u,v)h+\mathcal{o}(h)\\ &y_2=y(u+h,v)=y(u,v)+y_u(u,v)h+\mathcal{o}(h) \end{align*} $$
$$ \begin{align*} M_3\colon&x_3=x(u+h,v+h)=x(u,v)+x_u(u,v)h+x_v(u,v)h+\mathcal{o}(h)\\ &y_3=y(u+h,v+h)=y(u,v)+y_u(u,v)h+y_v(u,v)h+\mathcal{o}(h) \end{align*} $$
$$ \begin{align*} M_4\colon&x_4=x(u,v+h)=x(u,v)+x_v(u,v)h+\mathcal{o}(h)\\ &y_4=y(u,v+h)=y(u,v)+y_v(u,v)h+\mathcal{o}(h) \end{align*} $$
∵x1 + x3 = x2 + x4, y1 + y3 = y2 + y4
∴ M1M2M3M4为平行四边形
$$ \Rightarrow \Delta \sigma=\begin{vmatrix} x_2-x_1&y_2-y_1\\ x_4-x_1&y_4-y_1 \end{vmatrix}=\begin{vmatrix} x_u(u,v)h&y_u(u,v)h\\ x_v(u,v)h&y_v(u,v)h \end{vmatrix}=h^2\bigg|\frac{\partial(x,y)}{\partial(u,v)}\bigg| $$
$$ \Rightarrow \Delta\sigma=\bigg|\frac{\partial (x,y)}{\partial(u,v)}\bigg|\Delta \sigma^{'}+\mathcal{o}(\Delta \sigma^{'}) $$
$$ \therefore \iint_{D}f(x,y)\mathrm{d}\sigma=\iint_{D'}f(x(u,v),y(u,v))\bigg|\frac{\partial(x,y)}{\partial(u,v)}\bigg|\mathrm{d}u\mathrm{d}v $$
有用的定理
∬Df(x, y)dσ = ∬Df1(x)f2(y)dxdy = ∫abf1(x)dx∫cdf2(y)dy
三重积分
定义
$$ \iiint_{\Omega}f(x,y,z)\mathrm{d}V = \lim_{\lambda \to 0}\sum_{i=1}^{n}f(\xi_i, \eta_i,\zeta_i)\Delta V_i $$
计算
直角坐标
$$ \begin{align*} \iiint_{\Omega}f(x,y,z)\mathrm{d}V&=\int^{x_1}_{x_2}\mathrm{d}x\int^{y_1(x)}_{y_2(x)}\mathrm{d}y\int^{z_1(x,y)}_{z_2(x,y)}f(x,y,z)\mathrm{d}z\\ &=\int^{z_1}_{z_2}\mathrm{d}z\iint_{D_z}f(x,y,z)\mathrm{d}x\mathrm{d}y \end{align*} $$
柱面坐标
∭Ωf(x, y, z)dxdydz = ∭Ωf(ρcos θ, ρsin θ, z)ρdρdθdz
球面坐标
∭Ωf(x, y, z)dxdydz = ∭Ωf(rsin φcos θ, rsin φsin θ, rcos φ)r2sin φdrdφdθ
有用的定理
∭Ωf1(x)f2(y)f3(z)dxdydz = ∫abf1(x)dx∫cdf2(y)dy∫eff3(z)dz
重积分应用
曲面面积
$$ \begin{align*} A&=\iint_{D_{xy}}\sqrt{1+(\frac{\partial z}{\partial x})^2+(\frac{\partial z}{\partial y})^2}\mathrm{d}x\mathrm{d}y\\ &=\iint_{D_{xz}}\sqrt{1+(\frac{\partial y}{\partial x})^2+(\frac{\partial y}{\partial z})^2}\mathrm{d}x\mathrm{d}z\\ &=\iint_{D_{yz}}\sqrt{1+(\frac{\partial x}{\partial y})^2+(\frac{\partial x}{\partial z})^2}\mathrm{d}y\mathrm{d}z \end{align*} $$
质心
$$ 一般情况\colon\bar{x}=\frac{\iint_{D}x\mu(x,y)\mathrm{d}\sigma}{\iint_{D}\mu(x,y)\mathrm{d}\sigma},\;\bar{y}=\frac{\iint_{D}y\mu(x,y)\mathrm{d}\sigma}{\iint_{D}\mu(x,y)\mathrm{d}\sigma} $$
$$ 质量均匀\colon \bar{x}=\frac{1}{A}\iint_{D}x\mathrm{d}\sigma,\;\bar{y}=\frac{1}{A}\iint_{D}y\mathrm{d}\sigma $$
万有引力
$$ 三维\colon\begin{align*} \boldsymbol{F}&=(F_x,F_y,F_z)\\ &=(\iiint_{\Omega}G\frac{m\rho(x_0,y_0,z_0)(x-x_0)}{r^3}\mathrm{d}V,\iiint_{\Omega}G\frac{m\rho(x_0,y_0,z_0)(y-y_0)}{r^3}\mathrm{d}V,\iiint_{\Omega}G\frac{m\rho(x_0,y_0,z_0)(z-z_0)}{r^3}\mathrm{d}V) \end{align*} $$
$$ 二维\colon\begin{align*} \boldsymbol{F}&=(F_x, F_y, F_z)\\ &=(\iint_{D}G\frac{m\mu(x_0,y_0)(x-x_0)}{r^3}\mathrm{d}\sigma, \iint_{D}G\frac{m\mu(x_0,y_0)(y-y_0)}{r^3}\mathrm{d}\sigma, \iint_{D}G\frac{m\mu(x_0,y_0)(z-z_0)}{r^3}\mathrm{d}\sigma) \end{align*} $$